Rui is a professional deep water free diver. His altitude (in meters relative to sea level), $x$ seconds after diving, is modeled by: $d(x)=\dfrac{1}{2}x^2 -10x$ What is the lowest altitude Rui will reach?
Rui's altitude is modeled by a quadratic function, whose graph is a parabola. The lowest altitude is reached at the vertex. So in order to find the lowest altitude, we need to find the vertex's $y$ -coordinate. We will start by finding the vertex's $x$ -coordinate, and then plug that into $d(x)$. The vertex's $x$ -coordinate is the average of the two zeros, so let's find those first. $\begin{aligned} d(x)&=0 \\\\ \dfrac12x^2 -10x&=0 \\\\ x^2-20x&=0 \\\\ x(x-20)&=0 \\\\ \swarrow &\searrow \\\\ x=0\text{ or }&x-20=0 \\\\ x={0}\text{ or }&x={20} \end{aligned}$ Now let's take the zeros' average: $\dfrac{({0})+({20})}{2}=\dfrac{20}{2}={10}$ The vertex's $x$ -coordinate is $ {10}$. Now let's find $d({10})$ : $\begin{aligned} d({10})&=\dfrac12({10})^2-10({10}) \\\\ &=50-100 \\\\ &=-50 \end{aligned}$ In conclusion, the lowest altitude that Rui will reach is $-50$ meters relative to sea level.